Question: $f(n) = 6n+4-5(h(n))$ $h(t) = -6t-2$ $g(t) = -4t^{3}-4t^{2}-t-7-h(t)$ $ f(g(-2)) = {?} $
Solution: First, let's solve for the value of the inner function, $g(-2)$ . Then we'll know what to plug into the outer function. $g(-2) = -4(-2)^{3}-4(-2)^{2}-(-2)-7-h(-2)$ To solve for the value of $g$ , we need to solve for the value of $h(-2)$ $h(-2) = (-6)(-2)-2$ $h(-2) = 10$ That means $g(-2) = -4(-2)^{3}-4(-2)^{2}-(-2)-7-10$ $g(-2) = 1$ Now we know that $g(-2) = 1$ . Let's solve for $f(g(-2))$ , which is $f(1)$ $f(1) = (6)(1)+4-5(h(1))$ To solve for the value of $f$ , we need to solve for the value of $h(1)$ $h(1) = (-6)(1)-2$ $h(1) = -8$ That means $f(1) = (6)(1)+4+(-5)(-8)$ $f(1) = 50$